Question

Q1. I have two coins in my pocket, one is a fair coin, the other has heads on both sides. I pick one at random, and without looking at what it is, I toss it four times. I get four heads. (HHHH).

(I) What is the probability that I picked the fair coin?

(II) What is your answer if I got N heads in a row rather than four?

Q2. A, B are cowboys. A hits every shot. A has a ⅓ probability to hit his target on any shot. B has a ½ probability of hitting his target on any shot. A is to shoot first, and they will keep shooting until someone is hit.

(I) What is the probability that A survives the duel?

(II) What is the expected number of shots taken in this duel?

Q3. I only have a fair coin. Devise a procedure to pick one of three choices (A, B, C) so each choice is equally likely, using only the flips of the coin as the source of randomness. (Is there a non-repetitive way of doing this? If not, why not?)

Q4. I have two children.

(I) At least one of them is a girl. What is the probability that both are girls?

(II) The older one is a girl. What is the probability that both are girls?

(III) I have a daughter whose name is Jane. What is the probability that both are girls?

Answer 1

1) bayes theorem would be used in this question

P(chosing fair coin ) = 0.5

P(chosing non-fair coin ) = 0.5

P(heads coming from fair coin) = 0.5

P(heads from non-fair coin ) = 1

P(heads coming from fair coin in 4 trials ) = 0.5*0.5*0.5*0.5 = 0.0625

P(heads from non-fair coin in 4 trials ) = 1*1*1*1 = 1

so , using bayes theorem , P(getting HHHH from fair coin chosen) = .0625 * 0.5 / (0.0625*0.5 + 1* 0.5) = 1/17

2) If one would get N heads in a row then N being a large number probability of fair coin chosen would reduce as with increasing N probability of chosing a non-fair coin will tend towards 1 .